StringOutOfBoundsException 1 Java

Question

This is what the error says:

java.lang.StringIndexOutOfBoundsException: String index out of range: 1
    at java.lang.String.substring(String.java:1963)
    at FracCalc.produceAnswer(FracCalc.java:34)
    at FracCalc.main(FracCalc.java:16)

And Here is my existing code:

import java.util.*; 

public class FracCalc {

    public static void main(String[] args) 
    {
        // TODO: Read the input from the user and call produceAnswer with an equation
        Scanner input = new Scanner(System.in);
        System.out.println("Do you want to calculate something? If no, type in 'quit'. If yes, simply type in 'yes'. ");
        String Continue = input.next();
        if(Continue.equals("quit")){
            System.out.println("Have a nice day! ");
        }else if(Continue.equals("yes")){
        System.out.println("Please input a fraction, there must be exactly one space between the operator and the operand. ");
        String readInput = input.nextLine(); 
        System.out.println(produceAnswer(readInput)); 
        }
    }

// ** IMPORTANT ** DO NOT DELETE THIS FUNCTION.  This function will be used to test your code
// This function takes a String 'input' and produces the result
//
// input is a fraction string that needs to be evaluated.  For your program, this will be the user input.
//      e.g. input ==> "1/2 + 3/4"
//        
// The function should return the result of the fraction after it has been calculated
//      e.g. return ==> "1_1/4"
public static String produceAnswer(String input)
{ 
    // TODO: Implement this function to produce the solution to the input
    int space = input.indexOf(" ");
    int nextspace = space + 2;
    String operand = input.substring(0, space + 1);
    String operator = input.substring(space + 1, nextspace); 
    String operand2 = input.substring(nextspace + 1, input.length()); 
    return operand2;
}

// TODO: Fill in the space below with any helper methods that you think you will need

}

I'm trying to segregate fraction operators by determining the indexes of spaces and then going from there. Unfortunately i keep getting this error. Can someone please help! Thank You!

Possible duplicate of [Scanner is skipping nextLine() after using next(), nextInt() or other nextFoo()?](https://stackoverflow.com/questions/13102045/scanner-is-skipping-nextline-after-using-next-nextint-or-other-nextfoo) — Izruo, Nov 24 '17 at 00:22
I can sort of see how,but I still can't solve my problem. Would mind if you could kindly point out to what I should do on lines 34 and 16? — Maanas, Nov 24 '17 at 00:53

TheQuestioner · Accepted Answer · 2017-11-24T01:41:59.667

0

You have to use input.nextLine() since you want to read the entire line and not just characters. Try this in your main method:

    Scanner input = new Scanner(System.in);

    System.out.println("Do you want to calculate something? If no, type in 'quit'. If yes, simply type in 'yes'. ");
    String Continue = input.nextLine();
    if (Continue.equals("quit")) {
        System.out.println("Have a nice day! ");
    } else if (Continue.equals("yes")) {


        System.out.println(
                "Please input a fraction, there must be exactly one space between the operator and the operand. ");
        String readInput = input.nextLine();
        System.out.println(produceAnswer(readInput));
    }

edited Nov 24 '17 at 01:41

answered Nov 24 '17 at 01:15

TheQuestioner

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Hi, thank you so much - it did work! However, would you please explain to me what this does to my program? I was a bit puzzled at how it was formatted and initially thought it was a error on your end, but apparently not. And also why is it written in the middle of the line? – Maanas Nov 24 '17 at 01:29
I improved my answer. Try input.nextLine() and forget about the .toString(). It was written in the middle of the line because I use autoformat (if you use Eclipse, just hit ctrl +shift+f for autoformat your code). You can add your own StyleChecks aswell. – TheQuestioner Nov 24 '17 at 01:44
And to answer your question about the toString() method. It will return the String representation of this Scanner. – TheQuestioner Nov 24 '17 at 02:11

StringOutOfBoundsException 1 Java

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